After a totally inelastic collision, two objects of the same mass and same initial speeds are found to move together at half of their initial speeds. The angle between the initial velocities of the objects is

Options

A. `30^(@)`
B. `60^(@)`
C. `45^(@)`
D. `37^(@)`

Explanation

To solve the problem, we will use the principle of conservation of momentum. Let's break down the solution step by step.

### Step 1: Understand the Problem
We have two objects of the same mass \( m \) and the same initial speed \( V \). After a totally inelastic collision, they move together with a speed of \( \frac{V}{2} \). We need to find the angle \( \theta \) between their initial velocities.

**Hint**: Remember that in a totally inelastic collision, the two objects stick together after the collision.

### Step 2: Set Up the Momentum Conservation Equation
Before the collision, the momentum of each object can be expressed in vector form. Let’s denote the velocities of the two objects before the collision as:
- Object 1: \( \vec{v_1} = V \hat{i} \) (moving along the x-axis)
- Object 2: \( \vec{v_2} = V (\cos \theta \hat{i} + \sin \theta \hat{j}) \) (moving at an angle \( \theta \))

The total initial momentum \( \vec{P_{initial}} \) is:
\[ \vec{P_{initial}} = m \vec{v_1} + m \vec{v_2} = mV \hat{i} + mV (\cos \theta \hat{i} + \sin \theta \hat{j}) = mV(1 + \cos \theta) \hat{i} + mV \sin \theta \hat{j} \]

**Hint**: Break down the velocities into their x and y components.

### Step 3: Calculate Final Momentum
After the collision, the two objects move together with a speed of \( \frac{V}{2} \). The final momentum \( \vec{P_{final}} \) is:
\[ \vec{P_{final}} = 2m \left(\frac{V}{2}\right) \hat{v_f} = mV \hat{v_f} \]
where \( \hat{v_f} \) is the direction of their combined velocity after the collision.

**Hint**: The direction of \( \hat{v_f} \) can be determined from the components of the initial momenta.

### Step 4: Apply Conservation of Momentum
Since momentum is conserved, we equate the initial and final momentum:
\[ mV(1 + \cos \theta) \hat{i} + mV \sin \theta \hat{j} = mV \hat{v_f} \]

The magnitude of the final velocity \( \hat{v_f} \) is \( \frac{V}{2} \), so we can express \( \hat{v_f} \) in terms of its components:
\[ \hat{v_f} = \frac{(1 + \cos \theta)}{2} \hat{i} + \frac{\sin \theta}{2} \hat{j} \]

**Hint**: Remember that the magnitude of a vector can be found using the Pythagorean theorem.

### Step 5: Equate Components
From the x-component:
\[ 1 + \cos \theta = 1 \quad \text{(since the final speed is } \frac{V}{2} \text{)} \]
From the y-component:
\[ \sin \theta = 0 \]

**Hint**: Use trigonometric identities to relate the components.

### Step 6: Solve for \( \theta \)
From the equation \( 1 + \cos \theta = 1 \), we have:
\[ \cos \theta = 0 \implies \theta = 90^\circ \]

However, we need to consider the angle formed by the two velocities. Since the two objects are moving at an angle \( \theta \), we can use the relation:
\[ \cos \frac{\theta}{2} = \frac{1}{2} \]
This gives:
\[ \frac{\theta}{2} = 60^\circ \implies \theta = 120^\circ \]

**Hint**: Remember that the angle between two vectors can be found using the cosine rule.

### Conclusion
The angle between the initial velocities of the two objects is \( \theta = 120^\circ \).